Consider aₙ+1} = \; {1}{1 + \; 1/aₙ for n = 1, 2, , 2008, 2009 where a₁ = 1. Find the value of a₁ a₂ + a₂ a₃ + a₃ a₄ + + a₂008} a₂009}

Correct Answer is : 20082009\; \; 2008/200920092008

Correct Answer is : 20082009\; \; 2008/200920092008

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Xavier Aptitude Test 2023 Paper

Question : 65 of 102

Marks: +1, -0

Consider

a_{n+1} = \; \frac{1}{1 + \; \frac{1}{a_n}}

n = 1, 2, \ldots, 2008, 2009

a_1 = 1

a_1 a_2 + a_2 a_3 +

a_3 a_4 + \ldots + a_{2008} a_{2009}

Solution:

Given that

a_1 = 1 \& a_{n+1} = \; \frac{1}{1 + \; \frac{1}{a_n}}

a_2 = \; \frac{1}{1 + \; \frac{1}{1}} = \; \frac{1}{2}, a_3 = \; \frac{1}{1 + \; \frac{1}{\left( \; \frac{1}{2} \right)}} = \; \frac{1}{3}, \ldots

This implies,

a_n = \; \frac{1}{n}

Required value

= a_1 a_2 + a_2 a_3 + \cdots + a_{2008} a_{2009}

\; = \; \frac{1}{1} \times \; \frac{1}{2} + \; \frac{1}{2} \times \; \frac{1}{3} + \cdots + \; \frac{1}{2008} \times \; \frac{1}{2009}

\; = \left( \; \frac{1}{1} - \; \frac{1}{2} \right) + \left( \; \frac{1}{2} - \; \frac{1}{3} \right) + \cdots + \left( \; \frac{1}{2008} - \; \frac{1}{2009} \right)

\; = 1 - \; \frac{1}{2009}

\; = \; \frac{2008}{2009}

The answer is option C.

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